Question

Using properties of definite integrals, evaluate:
${\int }_0^{\frac{\pi }{2}}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$

Answer 1

Let $={\int }_0^{\pi /2}\frac{\sin x-\cos x}{1+\sin x\cos x}dx$
$I={\int }_0^{\pi /2}\frac{\sin x}{1+\sin x\cos x}dx-{\int }_0^{\pi /2}\frac{\cos x}{1+\sin x\cos x}dx$
Here $a+b-x=\frac{\pi }{2}-x$, so using the property
${\int }_a^{b}f\left(x\right)dx={\int }_a^{b}f(a+b-x)dx$
$I={\int }_0^{\pi /2}\frac{\sin (\frac{\pi }{2}-x)}{1+\sin (\frac{\pi }{2}-x)\cos (\frac{\pi }{2}-x)}dx-{\int }_0^{\pi /2}\frac{\cos x}{1+\sin x\cos x}dx$
$I={\int }_0^{\pi /2}\frac{\cos x}{1+\cos x\sin x}dx-{\int }_0^{\pi /2}\frac{\cos x}{1+\sin x\cos x}$
$I=0$